What is log_{1/3}(3x+4)>log_{1/3}(x^2+2) ?

The solution to log_{1/3}(3x+4)>log_{1/3}(x^2+2) is -4/3 (sqrt(17)+3)/2

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log_{1/3}(3x+4)>log_{1/3}(x^2+2)

lo g_{\frac{1}{3}} (3 x + 4) > lo g_{\frac{1}{3}} (x^{2} + 2)

- \frac{4}{3} < x < \frac{- 17 + 3}{2} or x > \frac{17 + 3}{2}

Interval Notation

(- \frac{4}{3}, \frac{- 17 + 3}{2}) \cup (\frac{17 + 3}{2}, \infty)

Decimal Notation

- 1.33333 \dots < x < - 0.56155 \dots or x > 3.56155 \dots

Solution steps

lo g_{\frac{1}{3}} (3 x + 4) > lo g_{\frac{1}{3}} (x^{2} + 2)

Apply log rules

3 x + 4 < x^{2} + 2 and 3 x + 4 > 0

3 x + 4 < x^{2} + 2 and 3 x + 4 > 0 : - \frac{4}{3} < x < \frac{- 17 + 3}{2} or x > \frac{17 + 3}{2}

- \frac{4}{3} < x < \frac{- 17 + 3}{2} or x > \frac{17 + 3}{2}

lo g_{- 6 x - 5 x^{2}} (6^{x}) > 0

ln (x + 1) < x

lo g_{10} (x + 5) > 1

ln (x - 2) \geq - 3

lo g_{3} (x + 2) < 1

What is log_{1/3}(3x+4)>log_{1/3}(x^2+2) ?
The solution to log_{1/3}(3x+4)>log_{1/3}(x^2+2) is -4/3 <x<(-sqrt(17)+3)/2 \lor x>(sqrt(17)+3)/2